Solve for $x$ : $2x^2 + 32x + 128 = 0$
Dividing both sides by $2$ gives: $ x^2 + {16}x + {64} = 0 $ The coefficient on the $x$ term is $16$ and the constant term is $64$ , so we need to find two numbers that add up to $16$ and multiply to $64$ The number $8$ used twice satisfies both conditions: $ {8} + {8} = {16} $ $ {8} \times {8} = {64} $ So $(x + {8})^2 = 0$ $x + 8 = 0$ Thus, $x = -8$ is the solution.